00:01
Alright, this problem isn't difficult, but there is a lot to it.
00:04
There's five different parts to it.
00:06
We're gonna take this cubic function x to the third minus 11 x squared plus 55 x minus 125 and it asks how many zeros does f of x have? well, we see that the power is 3 so the fundamental theorem of algebra tells us we have 3 zeros.
00:27
Okay, now according to descartes rule of sines, how many positive zeros and negative zeros are there? alright, so descartes rule of sine is asking how many sine changes do we have? so if we notice this was positive then it went to negative so that was 1 then it went back to positive so that was 2 and then it went back to negative.
00:50
That was 3 so what that tells us is you have 3 sine changes.
00:55
So that means we have 3 or 1 positive real zeros and it's 3 or 1 for a reason.
01:10
Alright, it's always down by an even number so 3 or 1 positive real zeros, then if i want to check for negative real zeros, i have to plug in negative x so you could take your original problem and plug in a negative x and we want to see how many times the sine changes when x is negative.
01:34
So if i have a negative x and that's cubed that remains negative negative x squared is positive, but i have to multiply that by a negative.
01:44
So that's going to be a negative 11 x squared 55 times a negative x is a negative 55 x and the minus of 125 and if you look every one of these is negative, right? so how many sine changes? none, so we have zero sine changes.
02:02
So that indicates to us that we have zero negative real zeros.
02:13
All right, it says show your work i hope that's enough to show your work and it says what are the potential? rational zeros.
02:23
All right potential rational zeros so when you're talking about your possible rational roots, you're gonna do a ratio of the factors of your constant term divided by factors of your leading coefficient all right, so our constant term is negative 125 so that means that our numerator the factors of our constant term could be plus or minus 1 plus or minus 125 right 1 times 125 makes that and 5 and 25 so plus and minus 5 plus and minus 25 our leading coefficient is just 1 so that just is plus or minus 1 so all of the potential rational zeros well, they're plus and minus 1 plus and minus 5 plus and minus 25 plus and minus 125 all right, and they want you to find the real and complex zeros if they exist.
03:39
So if you can imagine this, i'm not sure how much work they want, but you may need to start with synthetic division and do one and do your synthetic division into your problem and make sure that you get a zero for your remainder that tells you that that value that you chose 1 or negative 1 5 or negative 5 25 or negative 25 whatever value you chose was an actual zero of the function so to save a little bit of time i'll just start with 5 i graphed it to look and 5 is an x -intercept.
04:17
So 5 is going to be a 0 so i'll pull out my leading coefficients from my original problem 1 negative 1155 and negative 125 i'll skip my line and start my synthetic division i bring the 1 down and multiply 1 times 5 is 5 you do what it says negative 11 plus 5 is negative 6 remultiply so negative 6 times 5 is negative 30 do what it says 55 minus 30 is 25 remultiply 25 times 5 is 125 and when you subtract negative 125 from 120 or jeez negative 125 plus 125 is 0.
04:59
All right, so that leads us to our depressed polynomial remember our depressed polynomial is always 1 degree less than what we started with and since our original problem was x cubed my depressed polynomial is x squared okay, so remember earlier how we said that we had we could have positives negatives or imaginary answers, right? they're also talking about complex roots well, if i had three positive real zeros like they were talking about and no negative real zeros...